Now we look at $\{x: x^3 = u\}$ for some $u$, and we want to calculate its size. This can be interpreted as a sum of cubic characters: $\chi(u) + \chi^2(u) + \chi^3(u)$, where $\chi$ is a cubic character on $\mathbb{Z}/p\mathbb{Z}$, i.e. $\chi \neq 1$, but $\chi^3 = 1$. Now where does this interpretation come from?

If $x^3 = u$ has one solution, it has 3 because $p \equiv 1 \pmod{3}$ tells us that 1 has 3 cube roots in $\mathbb{Z}/p\mathbb{Z}$. On the character side, if $u$ is a cube of something, using $\chi^3 = 1$ w see that $\chi(u) + \chi^2(u) + \chi^3(u) = 3$.

In our case,
$(*) = \sum_{c+d = 1, c,d \neq 0} \left(1 + \chi(c/a) + \chi^2(c/a)\right)\left(1 + \chi(-d/b) + \chi^2(-d/b)\right)$.
There are three types of sum in the expansion. I would do one in each type to show you how it feels like.

$\displaystyle \sum_{c+d = 1, c,d \neq 0} \chi(c/a) = -\chi(1/a)$. To show this, note that for a nontrivial character $\chi$, $\displaystyle \sum_{u \in \mathbb{Z}/p\mathbb{Z}} \chi(u) = 0$, by an argument similar to what we showed in step 2 in the last block.

and that if you check out the books I mentioned before, $J$ can be expressed by Gauss sums of $\chi$ or $\chi^2$.

Summary:

So after all these work, we can find $\displaystyle \sum_{\stackrel{c+d = 1}{c,d \neq 0}} \#\{x:ax^3 = c\} \times \#\{y: -by^3 = d\} $ in terms of Gauss sums. There remains the case $c = 0$ or $d = 0$, which is easy to do directly. Sum them up, and you get the answer. The two books I mentioned both have examples of these types, so you may want to read them.